Given triangle ABC with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence. Find $\alpha, \beta, \gamma$.

Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.

My attempts so far:
Let $c=b+k$ and $a=b+2k$. Then, from $\alpha=2\beta$, I got $\gamma = 180^\circ - 3\beta$.

With the Law of Sines , I can write $$\dfrac{b}{\sin \beta}=\dfrac{b+k}{\sin 3\beta}=\frac{b+2k}{\text{sin } 2\beta}$$

From $\dfrac{b}{\sin\beta}=\dfrac{b+k}{\sin 3\beta}$, it gives $\cos 2\beta = \dfrac{k}{2b}$.

From $\dfrac{b}{\sin \beta}=\dfrac{b+2k}{\sin 2\beta}$, it gives $\cos \beta = \dfrac{b+2k}{2b}$.

After this, I don't know what should I do. Is there any other theorem that can be used to solve this problem?

Solution 1:

From $b,c,a$ forming an arithmetic sequence, $$c-b=a-c$$ $$2c=a+b$$ Applying sine rule, $$2\sin C=\sin A+\sin B$$ Using the given data $A=2B$, we can rewrite the above as, $$2\sin 3B=\sin2B+\sin B$$ $$\require{cancel}2\cdot \cancel{2\sin\frac{3B}{2}}\cos\frac{3B}2=\cancel{2\sin\frac{3B}{2}}\cos\frac B2\qquad\small{(\because\sin\frac{3B}{2}\neq0})$$ Now letting $\cos\frac B2=x$, we get, $$2(4x^3-3x)=x$$ $$x^2=\frac78$$ Therefore, $$\cos B=2\cos^2\frac B2-1=\frac34$$ $$B=\cos^{-1}\frac34$$

So we have found $B$ alias $\beta$ which leads to find other angles as well.

Solution 2:

Lemma: Since $\angle A=2\angle B$, $$a^2=b(b+c)$$ Proof: In a triangle $∠A=2∠B$ iff $a^2=b(b+c)$

Using the above result, $$(c+k)^2=(c-k)(2c-k)\implies (c+k)^2=(c-k)^2+c(c-k)\implies 4ck=c(c-k) $$ $$\therefore\; c=5k$$ Hence, the sides of the triangle are $(6k,4k,5k)$ and you can apply the cosine rule to finish it off. $$(4k)^2=(5k)^2+(6k)^2-2\cdot(5k)\cdot (6k)\cos \beta$$ $$\therefore\; \beta =\cos^{-1}\left(\frac{3}{4}\right)$$